Weierstrass Preparation Theorem, simple exercise

I am working with a specific version of the Weierstrass preparation theorem, trying to understand it with an example. I'll cite the theorem before stating the question:

TheoremIf $f:U\subset \mathbb{C}^{n+1}\rightarrow \mathbb{C}$ is a holomorphic function in the variables $(z,w_1,\dots,w_n)$ such that $f(z,0,\dots,0)$ is not identically zero near $z=0$, then there exist

• a holomorphic function $h:B \subset \mathbb{C}^{n+1}\rightarrow \mathbb{C}$ with $h(0,\dots,0)\neq 0$,
• a Weierstrass polynomial of the form$$ p(z,w_1,\dots,w_n) = z^d + \alpha_1(w_1,\dots,w_n) z^{d-1} + \dots + \alpha_d(w_1,\dots,w_n) $$where the coefficients are holomorphic functions in $(w_1,\dots,w_n)$ vanishing at the origin,

such that locally near the origin in $\mathbb{C}^{n+1}$ we have $f=h\cdot p$. The polynomial $p$ is unique.

Question: I am trying to get the Weierstrass polynomial for the following function $f:\mathbb{C}\rightarrow \mathbb{C}$ given by

$$ f(z_1,z_2) = z_1^3z_2 + z_1 z_2 + z_1^2z_2² + z_2^2 + z_1 z_2^3 $$

Obviously $f(z_1,0)$ vanishes identically, so the variable that I need to work on is $z_2$. By differentiating $f=h\cdot p$ with respect to $z_2$ I have been able to show that the degree $d$ is necessarily 2, i.e.$$ p(z_1,z_2) = z_2^2+\alpha(z_1)z_2 + \beta(z_1), $$for holomorphic coefficients $\alpha,\beta$ vanishing at $z_1=0$. However, I am stuck since I have too many degrees of freedom: $h,\alpha,\beta$. How can I construct a method for finding this coefficients?

1 Answer

I have found a solution to the problem by reading the proof of the WPT (as in Proposition 1.1.6 of Huybrechts' Complex Geometry).

We first note that the sections $f(z_1,0) \equiv 0$ and $f(0,z_2)= z_2^2$ so we can only single out the coordinate $z_2$ as "the good coordinate". In order to construct the Weierstrass polynomial, the proof of theorem considers

$$ p_{z_1}(z_2) = \prod_{i=1}^{d} (z_2 - a_i(z_1)), $$where $a_i(z_1)$ are the roots of $f_{z_1}(z_2)$ that restrict to the original root $z_2=0$ when the parameter $z_1=0$. The proof of the theorem gives an argument that shows that this is a Weierstrass polynomial in which the coefficients $\alpha_j$ are holomorphic, and written in terms of sums of the form$$ \sum_{i=1}^{k}a_i(z_2)^k. $$which then proves that are holomorphic by using the residue theorem.

In my particular case I can calculate the zeros of $f_{z_1}(z_2)$ by the quadratic formula. Factoring out $z_2$:$$ f_{z_1}(z_2) = z_2(z_1^3 + z_1 + (z_1^2+z_1)z_2 + z_1 z_2^2) $$and by the quadratic formula I get\begin{align*} a_0(z_1) & = 0,\\ a_1(z_1) &= \frac{-1-z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1},\\ a_2(z_1) & = \frac{-1-z_1^2 - \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1}. \end{align*}It is easy to see that $a_2(z_1)$ is not well defined for $z_1\rightarrow 0$, and that $a_1(z_1)\rightarrow 0$ for $z_1\rightarrow 0$. We consider then the Weierstrass polynomial$$ p_{z_1}(z_2) = z_2 \left( z_2 - \frac{-1-z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2 z_1}\right). $$This implies that the holomorphic function $h(z_1,z_2)$ is precisely\begin{align*} h(z_1,z_2) & = f(z_1,z_2)/p_{z_1}(z_2) = \frac{z_1 z_2 (z_2-a_1(z_1))(z_2-a_2(z_1))}{z_2(z_2-a_1(z_2))} \\ & = z_1 (z_2 -a_2(z_1)) = z_1z_2 + \frac{1+z_1^2 + \sqrt{-3 z_1^4 - 2 z_1^2 + 1}}{2} \end{align*}which, note, does not vanish at the origin $h(0,0)= 1$ and is thus a unit in the ring of germs $\mathcal{O}_{\mathbb{C}^{n},0}$.